Respuesta :
The first step to solve this question is to state the neutralization reaction between KOH and HClO:
[tex]KOH+HClO\rightarrow KClO+H_2O[/tex]Now, find the amount of moles of HClO present in 50mL of 0.170M HClO:
[tex]50.0mL\cdot\frac{1L}{1000mL}\cdot\frac{0.170mol}{L}=0.0085mol[/tex]According to the equation, 1 mole of KOH reacts with 1 mole of HClO. Use this ratio to find the amount of moles of KOH that react with 0.0085moles of HClO:
[tex]0.0085molHClO\cdot\frac{1molKOH}{1molHClO}=0.0085molKOH[/tex]Multiply this amount of moles of KOH by the inverse of its conversation:
[tex]0.0085molKOH\cdot\frac{1L}{0.170molKOH}=0.05L[/tex]It means that 0.05L of 0.170M KOH are needed to neutralize 50mL of 0.170M HClO.
